# Distance Minimization

1. Sep 29, 2007

### fubag

Find the point P on the parabola y = x^2 closest to the point (1,0).

To solve for P...

I used the distance formula, and then took the derivate. Using points (x, x^2) and (1,0)...

however, while taking the derivative I get a really nasty 4th power equation, which I am not sure I can solve without a calculator.

2. Sep 29, 2007

### EnumaElish

Can you derive your solution here?

3. Sep 29, 2007

I don't know how you did it, but the big hint here is of course to minimize the distance *squared*, rather than the distance.

4. Sep 30, 2007

### fubag

ok this is what I have so far...

D = sqrt. ((x-2)^2 + (x^2)^2)

then I took the derivative...

dD/dx = (1/2)((x-1)^2 + x^4)^-1/2)*(2(x-1) + 4x^3)

this simplifies to..

dD/dx = ((x-1) +2x^3)/sqrt.((x-1)^2 + x^4)

setting that equal to zero...

this is where I encounter issues..

2x^3 + x - 1 =0

and

sqrt. ((x-1)^2 + x^4) = 0

5. Sep 30, 2007

### HallsofIvy

Staff Emeritus